将 Pandas 列中出现的频率映射到字典

Question

我有一个 Pandas 数据框 df，其中包含这些列 -

Quiz          Question         K
  1              1           XYZ-01
  1              2           XXY-07
  1              2           XXP-05
  2              1           PQR-02
  2              1           WXY-09
  2              2           PQR-09
  2              3           XXY-07
  2              3           XXP-05

我想制作一个字典，其中包含Ks 在同一测验中的所有唯一组合作为键，并将它们的出现频率作为值。例如。对于上述 df，dict_combinations 应该类似于 -

{('XYZ-01', 'XXY-07'): 2, ('XXY-07', 'XXP-05'): 2, ('XYZ-01', 'XXP-05'): 1,       # for Quiz 1
 ('PQR-02', 'WXY-09'): 1, ('PQR-02', 'PQR-09'): 1, ('PQR-02', 'XXY-07'): 1, ('PQR-02', 'XYZ-01'): 1, # for Quiz 2 
 ('WXY-09', 'PQR-09'): 1, ('WXY-09', 'XXY-07'): 1, ('WXY-09', 'XXP-05'): 1, ('PQR-09', 'XXY-07'): 1,
 ('PQR-09', 'XXP-05'): 1}

我该怎么做？

Answer 1

使用DataFrame.dropna 删除可能的缺失值，然后使用DataFrame.sort_values 并在lambda 函数中为每个组创建2 的组合，通过Series.explode 展平嵌套元组，最后通过Series.value_counts 计算并转换为字典：

from  itertools import combinations

d = (df.dropna(subset=['K'])
        .sort_values(['Quiz','K'])
        .groupby('Quiz')['K'].apply(lambda x: list(combinations(x, 2)))
        .explode()
        .value_counts()
        .to_dict()
        )
print (d)
{('XXP-05', 'XXY-07'): 2, ('WXY-09', 'XXY-07'): 1, ('PQR-02', 'PQR-09'): 1,
 ('PQR-09', 'WXY-09'): 1, ('XXY-07', 'XYZ-01'): 1, ('PQR-02', 'XXY-07'): 1,
 ('PQR-09', 'XXP-05'): 1, ('PQR-02', 'WXY-09'): 1, ('PQR-09', 'XXY-07'): 1, 
 ('WXY-09', 'XXP-05'): 1, ('PQR-02', 'XXP-05'): 1, ('XXP-05', 'XYZ-01'): 1}

另一个想法是使用Counter 作为字典：

from  itertools import combinations
from collections import Counter

L = [tuple(sorted(y)) for i,x in df.dropna(subset=['K']).groupby('Quiz') 
                      for y in combinations(x['K'], 2)]
d = dict(Counter(L))
print (d)
{('XXY-07', 'XYZ-01'): 1, ('XXP-05', 'XYZ-01'): 1, ('XXP-05', 'XXY-07'): 2, 
 ('PQR-02', 'WXY-09'): 1, ('PQR-02', 'PQR-09'): 1, ('PQR-02', 'XXY-07'): 1, 
 ('PQR-02', 'XXP-05'): 1, ('PQR-09', 'WXY-09'): 1, ('WXY-09', 'XXY-07'): 1, 
 ('WXY-09', 'XXP-05'): 1, ('PQR-09', 'XXY-07'): 1, ('PQR-09', 'XXP-05'): 1}

编辑：explode 工作原理：

它扁平化Series 中的嵌套元组列表：

s = (df.dropna(subset=['K'])
        .sort_values(['Quiz','K'])
        .groupby('Quiz')['K'].apply(lambda x: list(combinations(x, 2)))
        # .explode()
        # .value_counts()
        # .to_dict()
        )
print (s)
Quiz
1    [(XXP-05, XXY-07), (XXP-05, XYZ-01), (XXY-07, ...
2    [(PQR-02, PQR-09), (PQR-02, WXY-09), (PQR-02, ...
Name: K, dtype: object

---

s = (df.dropna(subset=['K'])
        .sort_values(['Quiz','K'])
        .groupby('Quiz')['K'].apply(lambda x: list(combinations(x, 2)))
        .explode()
        )
print (s)
Quiz
1    (XXP-05, XXY-07)
1    (XXP-05, XYZ-01)
1    (XXY-07, XYZ-01)
2    (PQR-02, PQR-09)
2    (PQR-02, WXY-09)
2    (PQR-02, XXP-05)
2    (PQR-02, XXY-07)
2    (PQR-09, WXY-09)
2    (PQR-09, XXP-05)
2    (PQR-09, XXY-07)
2    (WXY-09, XXP-05)
2    (WXY-09, XXY-07)
2    (XXP-05, XXY-07)
Name: K, dtype: object